Arrange 1 / 2xy-4x ^ 2Y ^ 3 + 4Y ^ 2-x ^ 3 by ascending power

Arrange 1 / 2xy-4x ^ 2Y ^ 3 + 4Y ^ 2-x ^ 3 by ascending power

1/2xy+4y^2-x^3-4x^2y^3

Given 1 / x + 1 / y = 4, find the value of 4x-2xy + 4Y / x + 3xy + y

4x-2xy+4y/x+3xy+y
Where the numerator and denominator are given XY at the same time
=[4/y-2+4/x]/[1/y+3+1/x]
=[4(1/x+1/y)-2]/[(1/x+1/y)+3]
=(4×4-2)/(4+3)
=14/7
=2

Given 1 / X-1 / y = 5, find the value of (2Y + 3xy-2x) / (- 4x + 2XY + 4Y)

(2y+3xy-2x)/(-4x+2xy+4y)
=The denominator of (2 / x + 3-2 / x) / (- 4 / y + 2 + 4 / x) is also divided by XY
=[-2(1/x-1/y)+3]/[4(1/x-1/y)+2]
=(-10+3)/(20+2)
=-7/22

The equation 4 * + (a + 4) 2 * + 1 = 0 of X has a solution. To find the value range of real number a, the asterisk is x power

Because 2 * > 0
So let m = 2*
Then M & # 178; + (a + 4) m + 1 = 0 has a positive heel
If there is a root, then (a + 4) & # - 4 > = 0
a+4=2
a=2
If you don't have a direct follow
Because m1m2 = 1 > 0
Then both are negative
So M1 + M2 = - (a + 4) - 4
So if you have a positive heel, then a

The domain of the function f (x) = logx + 1 (5-x) is_______ Note: x + 1 is the base

(- 1,0) and (0,5)

Find the maximum and minimum value of the function y = (log (1 / 2) x) - (1 / 2log1 / 2x) + 5 in the interval [2,4]

2=log(1/2)4
Let a = log (1 / 2) x
That is - 2

Find the maximum and minimum values of the function y = (log2, x) ^ 2-2log2 (x ^ 2) + 6) on X ∈ [1, a]

y=(1og2 x)^2-2log2 (x^2)+6
=(log2 x)^2-4log2 x +6
Let log2 x = t, then t ∈ [0,1og2 a]
y=t^2-4t+6
The axis of symmetry is t = 2
Drawing analysis:
If log2 a

F (x) = (10-A) X-5, x0, a ≠ 1

This is a piecewise function
Because f (x) is an increasing function on R, there is a
10-A > 0 (determine that f (x) = (10-A) X-5 is an increasing function)
a> 1 (determine f (x) = logx with a as the base and X ≥ 1 as the increasing function)
The limit of F (x) = (10-A) X-5 at 1 is less than the function value of F (x) = logx with a as the base and X ≥ 1 at 1
10-a-5

Function: F (x) = logx ^ (1 + x) / (1-x), (a > 0. A ≠ 1), when a > 1, find the value range of X that makes f (x) > 0-
Hungry. Sorry

f(x)=loga (1+x)/(1-x)?
f(x)=loga (1+x)/(1-x)>0
f(x)=loga (1+x)/(1-x)>loga 1
(1+x)/(1-x)>1
(1+x)/(1-x)-1>0
(1+x-1+x)/(1-x)>0
2x(x-1)

The known function f (x) = {(2-3a) X-1, X

The problem is solved in three parts
When X & gt; = 1, the value of a is obviously. 0 & lt; a & lt; 1
2. When 0 & lt; X & lt; 1, the solution of. 2-3a & gt; = 1 & nbsp; is a & lt; = 1 / 3
3. When X & lt; 0, the solution of. 2-3a & lt; 0 & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; is a & lt; 2 / 3
To sum up the above answers, choose D
The key to this problem is that f (x) always develops to positive infinity